\(x^3+3x=3x^2+1\)
\(\Leftrightarrow x^3+3x-3x^2-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
\(x^2-3x=x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge-\dfrac{9}{4}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)
Vậy Min của b/t trên là : \(-\dfrac{9}{4}\Leftrightarrow x=\dfrac{3}{2}\)